Liquid is poured from a jar from top of a building
h = 20m high into a bathtub on the ground.
If a loudspeaker in the bathtub 'beep's, how much time
will it take for the sound to propagate to a microphone
inside the jar?
http://alexandersemenov.tripod.com/water/index.htm
Speed of sound in liquid is s = 30 m/s.
Assume that sound propagates through liquid only.
2007-03-21 08:01:50 · 5 answers · asked by Alexander 6 in Science & Mathematics ➔ Physics
The velocity of the water stream (yes, the cross section will narrow as it has to, we'll assume smooth Bernoulli flow) varies as √(2g(20-h)) where h is measured from the ground. The net upward speed of sound in the water is then
v = 30 - √(2g(20-h))
The total time is found by the definite integral:
∫ dh/(30 - √(2g(20-h)) from 0 to 20, and it comes to 1.28256 s
2007-03-21 08:31:19 · answer #1 · answered by Scythian1950 7 · 2⤊ 0⤋
The sound will travel up the water at a constant velocity, 30 m/s. However, the water is falling under the influence of gravity; the sound will need to travel "upstream" against this motion. So you have
x = v0 * t - ½ g * t²
20 m = 30 m/s * t - ½ * 9.8m/s² * t²
Leaving the units out for the moment to ease writing, this simplifies to
4.9 t² - 30 t + 20 = 0
Use the quadratic equation to solve:
t = {30 ± √(30² - 4*4.9*20)}/2*4.9
t = 30 ± √(508)}/9.8
t = 0.76 seconds
2007-03-21 08:20:09 · answer #2 · answered by Grizzly B 3 · 0⤊ 0⤋
If the sound goes 30 meters per second, and the microphone is 20 meters away, then it should take 2/3 second.
2007-03-21 08:45:45 · answer #3 · answered by TitoBob 7 · 0⤊ 0⤋
It will never reach the top because there is no way to keep the stream continuous and unbroken over a 20 m drop.
2007-03-21 08:35:13 · answer #4 · answered by catarthur 6 · 1⤊ 0⤋
since d = vt ==> t = d/v = 20/30= 0.66 s
2007-03-21 08:06:38 · answer #5 · answered by physicist 4 · 0⤊ 0⤋